When a sine wave travels through a region, two particles A and B...

Let the equation of the traveling sine wave be $y = A \sin(\omega t - kx)$.

The phase of a particle at position $x$ is $\phi = \omega t - kx$. For particles $A$ and $B$ with a phase difference of $\pi$, we have $\phi_A - \phi_B = \pi$.

The displacements of the particles are:
$y_A = A \sin(\phi_A)$
$y_B = A \sin(\phi_B) = A \sin(\phi_A - \pi) = -A \sin(\phi_A) = -y_A$
Since $y_B = -y_A$, the magnitudes of their displacements are always equal ($|y_A| = |y_B|$).

The velocities of the particles are:
$v_A = \dfrac{dy_A}{dt} = A\omega \cos(\phi_A)$
$v_B = \dfrac{dy_B}{dt} = A\omega \cos(\phi_A - \pi) = -A\omega \cos(\phi_A) = -v_A$
Since $v_B = -v_A$, the particles always move in opposite directions.

Both particles are excited by the same wave, so they oscillate with the same frequency. Thus, the frequency of $A$ is equal to the frequency of $B$.

The phase difference is related to the path difference by $\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x$. For the particles to be in opposite phase (equivalent to a phase difference of $\pi$, $3\pi$, $5\pi$, etc.), the separation can be $\Delta x = \left(n + \dfrac{1}{2}\right)\lambda$, where $n = 0, 1, 2, \dots$. Therefore, the separation does not have to be exactly half of the wavelength; it can be any odd multiple of $\dfrac{\lambda}{2}$.